《Cracking the Coding Interview》——第6章：智力题——题目3

2014-03-20 00:48

题目：有3升的瓶子和5升的瓶子，只允许倒满、倒到满为止、或是泼光三种操作，怎么搞出4升水呢？

解法：如果A和B是互质的两个正整数，且A<B，令X=B-A，则(X%B, 2X%B, 3X%B, ..., BX%B)正好就是0~n-1的一个排列。也就是说，两瓶子的容积如果互质，就可以配出0~B之间的任意容积。至于怎么配，请看代码。

代码：

// 6.3 Given two jugs with size x and y, and a volume v between 0 and y inclusive. Show how you're gonna get that value.
// Note that you can only fill a jug, empty a jug or pour from one into another.
// x and y will be positive and relatively prime.
#include <cstdio>
using namespace std;

int gcd(int x, int y)
{
    return x == 0 ? y : gcd(y % x, x);
}

int main()
{
    int x, y;
    int vx, vy;
    int v;
    
    while (scanf("%d%d%d", &x, &y, &v) == 3 && (x > 0 && y > 0 && v > 0)) {
        if (x > y) {
            continue;
        }
        if (gcd(x, y) > 1) {
            continue;
        }
        
        vx = 0;
        vy = y;
        printf("[%d, %d]: fill y\n", vx, vy);
        while (vy != v) {
            if (vy < x) {
                vx = vy;
                vy = 0;
                printf("[%d, %d]: pour y into x\n", vx, vy);
                vy = y;
                printf("[%d, %d]: fill y\n", vx, vy);
                vy = vy - (x - vx);
                vx = x;
                printf("[%d, %d]: pour y into x\n", vx, vy);
                vx = 0;
                printf("[%d, %d]: empty x\n", vx, vy);
            } else {
                vy -= x;
                vx = x;
                printf("[%d, %d]: pour y into x\n", vx, vy);
                vx = 0;
                printf("[%d, %d]: empty x\n", vx, vy);
            }
        }
    }
    
    return 0;
}